Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int grid[105][105];
struct step {
  int rc, n;
};
int main() {
  int n, m;
  cin >> n >> m;
  int flag = true;
  if (n < m)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) scanf("%d", &grid[i][j]);
  else {
    flag = false;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) scanf("%d", &grid[j][i]);
    swap(n, m);
  }
  list<step> ans;
  bool psb = true;
  for (int i = 1; i <= n; i++) {
    while (count(grid[i] + 1, grid[i] + 1 + m, 0) == 0) {
      for (int j = 1; j <= m; j++) grid[i][j]--;
      step tmp;
      tmp.n = i;
      tmp.rc = 0;
      ans.push_back(tmp);
    }
  }
  for (int j = 1; j <= m && psb; j++) {
    while (psb) {
      int cnt = 0;
      for (int i = 1; i <= n; i++) {
        if (grid[i][j] == 0) cnt++;
      }
      if (cnt == 0) {
        for (int i = 1; i <= n; i++) {
          grid[i][j]--;
        }
        step tmp;
        tmp.n = j;
        tmp.rc = 1;
        ans.push_back(tmp);
      } else if (cnt < n) {
        psb = false;
      } else
        break;
    }
  }
  if (psb) {
    cout << ans.size() << endl;
    for (auto &e : ans)
      cout << ((!flag) ^ e.rc == 0 ? "row " : "col ") << e.n << endl;
  } else
    cout << -1 << endl;
}